Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 10 - Section 10.2 - Systems of Linear Equations in Several Variables - 10.2 Exercises - Page 698: 46

Answer

6 days for Factory -A, 2 days for Factory -B, and 3 days for Factory -C.

Work Step by Step

Step 1. Assume $x$ days are needed for Factory -A, $y$ days are needed for Factory -B, and $z$ days are needed for Factory -C. Step 2. With the total number of different appliances from the orders, we can setup the following equations: $8x+10y+14z=110, 16x+12y+10z=150,10x+18y+6z=114$, which can be simplified as $4x+5y+7z=55, 8x+6y+5z=75, 5x+9y+3z=57$ Step 3. To eliminate variable $x$, raise all the coefficients before $x$ to 40 by multiply the first equation with 10, second with 5, and third with 8, we get $40x+50y+70z=550, 40x+30y+25z=375, 40x+72y+24z=456$ Step 4. Subtract the middle equation from the first and third, we get $20y+45z=175, 42y-z=81$ or $z=42y-81$ Step 5. Simplify the first equation to $4y+9z=35$ and use the second to back-substitute, we get $4y+9(42y-81)=35$ which can be solved to get $y=2$, thus $z=42\times2-81=3$ and $x=\frac{55-5\times2-7\times3}{4}=6$ Step 6. We conclude that 6 days are needed for Factory -A, 2 days are needed for Factory -B, and 3 days are needed for Factory -C.
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